4

我需要获取每个方法的调用者的源位置。我正在尝试创建一个proc_macro_attribute来捕获位置并打印它。

#[proc_macro_attribute]
pub fn get_location(attr: TokenStream, item: TokenStream) -> TokenStream {
    // Get and print file!(), line!() of source
    // Should print line no. 11
    item
}
#[get_location]
fn add(x: u32, y: u32) -> u32 {
    x + y
}

fn main() {
    add(1, 5); // Line No. 11
}
4

2 回答 2

3

TL;博士

这是一个程序宏,它使用synquote执行您所描述的操作:

// print_caller_location/src/lib.rs

use proc_macro::TokenStream;
use quote::quote;
use syn::spanned::Spanned;

// Create a procedural attribute macro
//
// Notably, this must be placed alone in its own crate
#[proc_macro_attribute]
pub fn print_caller_location(_attr: TokenStream, item: TokenStream) -> TokenStream {
    // Parse the passed item as a function
    let func = syn::parse_macro_input!(item as syn::ItemFn);

    // Break the function down into its parts
    let syn::ItemFn {
        attrs,
        vis,
        sig,
        block,
    } = func;

    // Ensure that it isn't an `async fn`
    if let Some(async_token) = sig.asyncness {
        // Error out if so
        let error = syn::Error::new(
            async_token.span(),
            "async functions do not support caller tracking functionality
    help: consider returning `impl Future` instead",
        );

        return TokenStream::from(error.to_compile_error());
    }

    // Wrap body in a closure only if function doesn't already have #[track_caller]
    let block = if attrs.iter().any(|attr| attr.path.is_ident("track_caller")) {
        quote! { #block }
    } else {
        quote! {
            (move || #block)()
        }
    };

    // Extract function name for prettier output
    let name = format!("{}", sig.ident);

    // Generate the output, adding `#[track_caller]` as well as a `println!`
    let output = quote! {
        #[track_caller]
        #(#attrs)*
        #vis #sig {
            println!(
                "entering `fn {}`: called from `{}`",
                #name,
                ::core::panic::Location::caller()
            );
            #block
        }
    };

    // Convert the output from a `proc_macro2::TokenStream` to a `proc_macro::TokenStream`
    TokenStream::from(output)
}

确保将其放入板条箱并将这些行添加到其Cargo.toml

# print_caller_location/Cargo.toml

[lib]
proc-macro = true

[dependencies]
syn = {version = "1.0.16", features = ["full"]}
quote = "1.0.3"
proc-macro2 = "1.0.9"

深入解释

宏只能扩展为可以手动编写的代码。知道了这一点,我在这里看到两个问题:

  1. 如何编写一个跟踪调用者位置的函数?
  2. 如何编写创建此类函数的程序宏?

初次尝试

我们想要一个程序宏

  • 接受一个函数,
  • 标记它#[track_caller]
  • 并添加一行打印Location::caller.

例如,它会像这样转换一个函数:

fn foo() {
    // body of foo
}

进入

#[track_caller]
fn foo() {
    println!("{}", std::panic::Location::caller());
    // body of foo
}

下面,我展示了一个完全执行该转换的过程宏——尽管,正如您将在以后的版本中看到的,您可能想要一些不同的东西。要尝试此代码,就像之前在 TL;DR 部分中一样,将其放入自己的 crate 并将其依赖项添加到Cargo.toml.

// print_caller_location/src/lib.rs

use proc_macro::TokenStream;
use quote::quote;

// Create a procedural attribute macro
//
// Notably, this must be placed alone in its own crate
#[proc_macro_attribute]
pub fn print_caller_location(_attr: TokenStream, item: TokenStream) -> TokenStream {
    // Parse the passed item as a function
    let func = syn::parse_macro_input!(item as syn::ItemFn);

    // Break the function down into its parts
    let syn::ItemFn {
        attrs,
        vis,
        sig,
        block,
    } = func;

    // Extract function name for prettier output
    let name = format!("{}", sig.ident);

    // Generate the output, adding `#[track_caller]` as well as a `println!`
    let output = quote! {
        #[track_caller]
        #(#attrs)*
        #vis #sig {
            println!(
                "entering `fn {}`: called from `{}`",
                #name,
                ::core::panic::Location::caller()
            );
            #block
        }
    };

    // Convert the output from a `proc_macro2::TokenStream` to a `proc_macro::TokenStream`
    TokenStream::from(output)
}

示例用法:

// example1/src/main.rs

#![feature(track_caller)]

#[print_caller_location::print_caller_location]
fn add(x: u32, y: u32) -> u32 {
    x + y
}

fn main() {
    add(1, 5); // entering `fn add`: called from `example1/src/main.rs:11:5`
    add(1, 5); // entering `fn add`: called from `example1/src/main.rs:12:5`
}

不幸的是,我们无法摆脱那个简单的版本。该版本至少存在两个问题:

  • 它如何与async fns 组成:

    • 它不是打印调用者位置,而是打印调用我们的宏 ( #[print_caller_location]) 的位置。例如:

    // example2/src/main.rs
    
    #![feature(track_caller)]
    
    #[print_caller_location::print_caller_location]
    async fn foo() {}
    
    fn main() {
        let future = foo();
        // ^ oops! prints nothing
        futures::executor::block_on(future);
        // ^ oops! prints "entering `fn foo`: called from `example2/src/main.rs:5:1`"
        let future = foo();
        // ^ oops! prints nothing
        futures::executor::block_on(future);
        // ^ oops! prints "entering `fn foo`: called from `example2/src/main.rs:5:1`"
    }
    
  • 它如何与自身的其他调用一起工作,或者一般来说,#[track_caller]

    • 嵌套函数#[print_caller_location]将打印根调用者的位置,而不是给定函数的直接调用者。例如:

    // example3/src/main.rs
    
    #![feature(track_caller)]
    
    #[print_caller_location::print_caller_location]
    fn add(x: u32, y: u32) -> u32 {
        x + y
    }
    
    #[print_caller_location::print_caller_location]
    fn add_outer(x: u32, y: u32) -> u32 {
        add(x, y)
        // ^ we would expect "entering `fn add`: called from `example3/src/main.rs:12:5`"
    }
    
    fn main() {
        add(1, 5);
        // ^ "entering `fn add`: called from `example3/src/main.rs:17:5`"
        add(1, 5);
        // ^ "entering `fn add`: called from `example3/src/main.rs:19:5`"
        add_outer(1, 5);
        // ^ "entering `fn add_outer`: called from `example3/src/main.rs:21:5`"
        // ^ oops! "entering `fn add`: called from `example3/src/main.rs:21:5`"
        //
        // In reality, `add` was called on line 12, from within the body of `add_outer`
        add_outer(1, 5);
        // ^ "entering `fn add_outer`: called from `example3/src/main.rs:26:5`"
        // oops! ^ entering `fn add`: called from `example3/src/main.rs:26:5`
        //
        // In reality, `add` was called on line 12, from within the body of `add_outer`
    }
    

寻址async fn_

async fn使用s可以解决这个问题-> impl Future,例如,如果我们希望我们的async fn反例正常工作,我们可以改为:

// example4/src/main.rs

#![feature(track_caller)]

use std::future::Future;

#[print_caller_location::print_caller_location]
fn foo() -> impl Future<Output = ()> {
    async move {
        // body of foo
    }
}

fn main() {
    let future = foo();
    // ^ prints "entering `fn foo`: called from `example4/src/main.rs:15:18`"
    futures::executor::block_on(future);
    // ^ prints nothing
    let future = foo();
    // ^ prints "entering `fn foo`: called from `example4/src/main.rs:19:18`"
    futures::executor::block_on(future);
    // ^ prints nothing
}

我们可以添加一个特殊情况,将这种转换应用于我们的宏。但是,除了影响返回的未来可能具有的自动特征之外,该转换还将函数的公共 API 从 更改为async fn foo()fn foo() -> impl Future<Output = ()>

因此,我建议我们允许用户根据需要使用该解决方法,如果我们的宏用于async fn. 我们可以通过将这些行添加到我们的宏代码中来做到这一点:

// Ensure that it isn't an `async fn`
if let Some(async_token) = sig.asyncness {
    // Error out if so
    let error = syn::Error::new(
        async_token.span(),
        "async functions do not support caller tracking functionality
    help: consider returning `impl Future` instead",
    );

    return TokenStream::from(error.to_compile_error());
}

#[print_caller_location]修复函数的嵌套行为

有问题的行为可以归结为以下事实:当一个#[track_caller]函数foo, 直接调用另一个#[track_caller]函数 ,bar时,Location::caller将使它们都可以访问foo的调用者。换句话说,在嵌套函数Location::caller的情况下可以访问根调用者:#[track_caller]

#![feature(track_caller)]

fn main() {
    foo(); // prints `src/main.rs:4:5` instead of the line number in `foo`
}

#[track_caller]
fn foo() {
   bar();
}

#[track_caller]
fn bar() {
    println!("{}", std::panic::Location::caller());
}

游乐场链接

为了解决这个问题,我们需要打破#[track_caller]调用链。我们可以通过将嵌套调用隐藏bar在闭包中来打破链条:

#![feature(track_caller)]

fn main() {
    foo();
}

#[track_caller]
fn foo() {
    (move || {
        bar(); // prints `src/main.rs:10:9`
    })()
}

#[track_caller]
fn bar() {
    println!("{}", std::panic::Location::caller());
}

游乐场链接

现在我们知道如何打破#[track_caller]函数链,我们可以解决这个问题。我们只需要确保如果用户真的#[track_caller]有意标记他们的函数,我们就不会插入闭包并破坏链。

我们可以将这些行添加到我们的解决方案中:

// Wrap body in a closure only if function doesn't already have #[track_caller]
let block = if attrs.iter().any(|attr| attr.path.is_ident("track_caller")) {
    quote! { #block }
} else {
    quote! {
        (move || #block)()
    }
};

最终解决方案

在这两个更改之后,我们最终得到了以下代码:

// print_caller_location/src/lib.rs

use proc_macro::TokenStream;
use quote::quote;
use syn::spanned::Spanned;

// Create a procedural attribute macro
//
// Notably, this must be placed alone in its own crate
#[proc_macro_attribute]
pub fn print_caller_location(_attr: TokenStream, item: TokenStream) -> TokenStream {
    // Parse the passed item as a function
    let func = syn::parse_macro_input!(item as syn::ItemFn);

    // Break the function down into its parts
    let syn::ItemFn {
        attrs,
        vis,
        sig,
        block,
    } = func;

    // Ensure that it isn't an `async fn`
    if let Some(async_token) = sig.asyncness {
        // Error out if so
        let error = syn::Error::new(
            async_token.span(),
            "async functions do not support caller tracking functionality
    help: consider returning `impl Future` instead",
        );

        return TokenStream::from(error.to_compile_error());
    }

    // Wrap body in a closure only if function doesn't already have #[track_caller]
    let block = if attrs.iter().any(|attr| attr.path.is_ident("track_caller")) {
        quote! { #block }
    } else {
        quote! {
            (move || #block)()
        }
    };

    // Extract function name for prettier output
    let name = format!("{}", sig.ident);

    // Generate the output, adding `#[track_caller]` as well as a `println!`
    let output = quote! {
        #[track_caller]
        #(#attrs)*
        #vis #sig {
            println!(
                "entering `fn {}`: called from `{}`",
                #name,
                ::core::panic::Location::caller()
            );
            #block
        }
    };

    // Convert the output from a `proc_macro2::TokenStream` to a `proc_macro::TokenStream`
    TokenStream::from(output)
}
于 2020-03-18T01:20:34.937 回答
2

可以使用现成的解决方案(请参阅@timotree 的评论)。如果您想自己执行此操作,具有更大的灵活性或学习,您可以编写一个程序宏来解析回溯(从被调用的函数内部获得)并打印您需要的信息。这是 a 中的一个过程宏lib.rs

extern crate proc_macro;
use proc_macro::{TokenStream, TokenTree};

#[proc_macro_attribute]
pub fn get_location(_attr: TokenStream, item: TokenStream) -> TokenStream {

    // prefix code to be added to the function's body
    let mut prefix: TokenStream = "
        // find earliest symbol in source file using backtrace
        let ps = Backtrace::new().frames().iter()
            .flat_map(BacktraceFrame::symbols)
            .skip_while(|s| s.filename()
                .map(|p|!p.ends_with(file!())).unwrap_or(true))
            .nth(1 as usize).unwrap();

        println!(\"Called from {:?} at line {:?}\",
            ps.filename().unwrap(), ps.lineno().unwrap());
    ".parse().unwrap(); // parse string into TokenStream

    item.into_iter().map(|tt| { // edit input TokenStream
        match tt { 
            TokenTree::Group(ref g) // match the function's body
                if g.delimiter() == proc_macro::Delimiter::Brace => { 

                    prefix.extend(g.stream()); // add parsed string

                    TokenTree::Group(proc_macro::Group::new(
                        proc_macro::Delimiter::Brace, prefix.clone()))
            },
            other => other, // else just forward TokenTree
        }
    }).collect()
} 

回溯被解析以找到源文件中最早的符号(使用file!()另一个宏检索)。我们需要添加到函数中的代码定义在一个字符串中,然后将其解析为 aTokenStream并添加到函数体的开头。我们本可以在最后添加这个逻辑,但是返回一个没有分号的值就不再起作用了。然后,您可以使用以下程序宏main.rs

extern crate backtrace;
use backtrace::{Backtrace, BacktraceFrame};
use mylib::get_location;

#[get_location]
fn add(x: u32, y: u32) -> u32 { x + y }

fn main() { 
    add(1, 41);
    add(41, 1);
}

输出是:

> Called from "src/main.rs" at line 10
> Called from "src/main.rs" at line 11

不要忘记lib通过将这两行添加到您的以下来指定您的板条箱提供程序宏Cargo.toml

[lib]
proc-macro = true
于 2020-03-18T00:11:28.937 回答